POJ2777--Count Color

Posted 2021-02-13 tanwei-hq

Description

Chosen Problem Solving and Program design as an optional course, you are required to solve all kinds of problems. Here, we get a new problem.

There is a very long board with length L centimeter, L is a positive integer, so we can evenly divide the board into L segments, and they are labeled by 1, 2, ... L from left to right, each is 1 centimeter long. Now we have to color the board - one segment with only one color. We can do following two operations on the board:

1. "C A B C" Color the board from segment A to segment B with color C.
2. "P A B" Output the number of different colors painted between segment A and segment B (including).

In our daily life, we have very few words to describe a color (red, green, blue, yellow…), so you may assume that the total number of different colors T is very small. To make it simple, we express the names of colors as color 1, color 2, ... color T. At the beginning, the board was painted in color 1. Now the rest of problem is left to your.

Input

First line of input contains L (1 <= L <= 100000), T (1 <= T <= 30) and O (1 <= O <= 100000). Here O denotes the number of operations. Following O lines, each contains "C A B C" or "P A B" (here A, B, C are integers, and A may be larger than B) as an operation defined previously.

Output

Ouput results of the output operation in order, each line contains a number.

Sample Input

2 2 4
C 1 1 2
P 1 2
C 2 2 2
P 1 2

Sample Output

2
1

题目描述：
首先对一个木板，长度为L，颜色种类为T，有O次询问，一开始建树，初始时将所有的树颜色设为1；
随后进行操作，如果改变颜色，改区间的颜色改变，包含此区间的颜色变为-1，因为这个区间不在是
纯色，这个区间有多种颜色。用一个颜色的数组储存所以颜色出现的次数，一开始都为0，查询时如果
查询的区间不为-1，表示该区间只有一种颜色，将这种颜色的数量+1，如果为-1.则继续搜索孩子。


代码：
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
#define maxn 100005
int colors[35],l,t,o,a,b,c,ans;
char ch;
struct node
{
 int left;
 int right;
 int color;
}tree[maxn<<2];
void build(int l,int r,int k)
{
 tree[k].left=l;
 tree[k].right=r;
 tree[k].color=1;
 if(l==r)
 return;
 int mid=(l+r)/2;
 build(l,mid,k<<1);
 build(mid+1,r,k<<1|1);
}
void update(int l,int r,int c,int k)
{
 if(tree[k].color==c)
 return;
 if(tree[k].left==l&&tree[k].right==r)
 {
  tree[k].color=c;
  return;
 }
 if(tree[k].color!=-1)
 {
  tree[k<<1].color=tree[k<<1|1].color=tree[k].color;
  tree[k].color=-1;
 }
 int mid=(tree[k].left+tree[k].right)/2;
 if(mid>=r)
 {
  update(l,r,c,k<<1);
 }
 else if(mid<l)
 {
  update(l,r,c,k<<1|1);
 }
 else
 {
  update(l,mid,c,k<<1);
  update(mid+1,r,c,k<<1|1);
 }
}
void qurey(int l,int r,int k)
{
 if(tree[k].color!=-1)
 {
  colors[tree[k].color]=1;
 }
 else
 {
  int mid=(tree[k].left+tree[k].right)/2;
  if(mid>=r)
  {
   qurey(l,r,k<<1);
  }
  else if(mid<l)
  {
   qurey(l,r,k<<1|1);
  }
  else
  {
   qurey(l,mid,k<<1);
   qurey(mid+1,r,k<<1|1);
  }
 }
}
int main()
{
 while(scanf("%d%d%d",&l,&t,&o)!=EOF)
 {
  build(1,l,1);
  while(o--)
  {
   getchar();
   scanf("%c",&ch);
   if(ch==‘C‘)
   {
    scanf("%d%d%d",&a,&b,&c);
    update(a,b,c,1);
   }
   if(ch==‘P‘)
   {
    memset(colors,0,sizeof(colors));
    scanf("%d%d",&a,&b);
    ans=0;
    qurey(a,b,1);
    for(int i=1;i<=t;i++)
    {
     if(colors[i]!=0)
     ans++;
    }
    printf("%d
",ans);
   }
  }
 }
 return 0;
}